Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.1$ years; the standard deviation is $0.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living between $1.3$ and $4.3$ years.
Answer: $3.1$ $2.5$ $3.7$ $1.9$ $4.3$ $1.3$ $4.9$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $3.1$ years. We know the standard deviation is $0.6$ years, so one standard deviation below the mean is $2.5$ years and one standard deviation above the mean is $3.7$ years. Two standard deviations below the mean is $1.9$ years and two standard deviations above the mean is $4.3$ years. Three standard deviations below the mean is $1.3$ years and three standard deviations above the mean is $4.9$ years. We are interested in the probability of a lizard living between $1.3$ and $4.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lizards will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of lizards between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular lizard living between $1.3$ and $4.3$ years is $\color{orange}{2.35\%} + {95\%}$, or $97.35\%$.